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postnew [5]
3 years ago
8

What is the constant of 2x+3y+6

Mathematics
2 answers:
STALIN [3.7K]3 years ago
6 0

Answer:

6

Step-by-step explanation:

vekshin13 years ago
3 0

Answer:

6

Step-by-step explanation:

Constants are any numbers that don't change.

2x + 3y + 6

Since "6" is the only number here, that is the only constant.

Best of Luck!

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The accompanying data on IQ for first-graders at auniversity lab school was introduced in Example 1.2.82 96 99 102 103 103 106 1
almond37 [142]

Answer:

Mean = 113.72

The estimator of 113.72 is point estimate used for the  conceptual population of all first-graders in this school

Step-by-step explanation:

Data provided in the question:

Data:

82 96 99 102 103 103 106 107 108 108 108 108 109 110 110 111 113 113 113 113 115 115 118 118 119 121 122 122 127 132 136 140 146

Number of data = 33

In the hint it is also given that the sum of the data value i.e ∑x = 3753

Now,

Point estimate of mean is given as;

Mean = \frac{\textup{Sum of all the data values}}{\textup{Total number of data value}}

or

⇒ Mean = \frac{\textup{3753}}{\textup{33}}

or

⇒ Mean = 113.72

Hence,

The estimator of 113.72 is point estimate used for the  conceptual population of all first-graders in this school

5 0
4 years ago
Subtract. <br> 23.87 - 1.12<br><br> A. 13.75<br><br> B. 22.75<br><br> C. 24.99<br><br> D. 12.67
Shtirlitz [24]

Answer:

B. 22.75

Step-by-step explanation:

23.87

<u>-   1.12</u>

22.75

7 0
3 years ago
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Which of the following is the best estimate of 31% of 15?<br> 7.5<br> 3<br> 5<br> 4
Alex777 [14]
Th answer is 5 because 31% of 15 is 4.65
3 0
3 years ago
Read 2 more answers
Find a power series for the function, centered at c, and determine the interval of convergence. f(x) = 9 3x + 2 , c = 6
san4es73 [151]

Answer:

\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ........

The interval of convergence is:(-\frac{2}{3},\frac{16}{3})

Step-by-step explanation:

Given

f(x)= \frac{9}{3x+ 2}

c = 6

The geometric series centered at c is of the form:

\frac{a}{1 - (r - c)} = \sum\limits^{\infty}_{n=0}a(r - c)^n, |r - c| < 1.

Where:

a \to first term

r - c \to common ratio

We have to write

f(x)= \frac{9}{3x+ 2}

In the following form:

\frac{a}{1 - r}

So, we have:

f(x)= \frac{9}{3x+ 2}

Rewrite as:

f(x) = \frac{9}{3x - 18 + 18 +2}

f(x) = \frac{9}{3x - 18 + 20}

Factorize

f(x) = \frac{1}{\frac{1}{9}(3x + 2)}

Open bracket

f(x) = \frac{1}{\frac{1}{3}x + \frac{2}{9}}

Rewrite as:

f(x) = \frac{1}{1- 1 + \frac{1}{3}x + \frac{2}{9}}

Collect like terms

f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2}{9}- 1}

Take LCM

f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2-9}{9}}

f(x) = \frac{1}{1 + \frac{1}{3}x - \frac{7}{9}}

So, we have:

f(x) = \frac{1}{1 -(- \frac{1}{3}x + \frac{7}{9})}

By comparison with: \frac{a}{1 - r}

a = 1

r = -\frac{1}{3}x + \frac{7}{9}

r = -\frac{1}{3}(x - \frac{7}{3})

At c = 6, we have:

r = -\frac{1}{3}(x - \frac{7}{3}+6-6)

Take LCM

r = -\frac{1}{3}(x + \frac{-7+18}{3}+6-6)

r = -\frac{1}{3}(x + \frac{11}{3}+6-6)

So, the power series becomes:

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}ar^n

Substitute 1 for a

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}1*r^n

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}r^n

Substitute the expression for r

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}(-\frac{1}{3}(x - \frac{7}{3}))^n

Expand

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}[(-\frac{1}{3})^n* (x - \frac{7}{3})^n]

Further expand:

\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ................

The power series converges when:

\frac{1}{3}|x - \frac{7}{3}| < 1

Multiply both sides by 3

|x - \frac{7}{3}|

Expand the absolute inequality

-3 < x - \frac{7}{3}

Solve for x

\frac{7}{3}  -3 < x

Take LCM

\frac{7-9}{3} < x

-\frac{2}{3} < x

The interval of convergence is:(-\frac{2}{3},\frac{16}{3})

6 0
3 years ago
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